where the second equality uses that \(\Phi (-x) \rightarrow -x^2/2\)  [1]} (Recall that \(\Phi (-x) \approx e^{-x^2/ 2}\)  for large \(x\) .)

It is clear from this result that adjusting the long-run sampling proportions based on the state of nature \(\theta \)  can result in faster learning. Define the optimal exponent and optimal long-run sampling ratios as
\(\Gamma ^{-1}_{\theta } \triangleq \max _{p\in \mathcal {S}} \min _{j\ne I^*} \frac{ \left(\mu (\theta , I^*, w) - \mu (\theta , j, w)\right)^2 }{2 \Vert  X_{\rm pop}\Vert _{A}^2 \left(p_i^{-1} + p_j^{-1}\right)} \qquad p^*(\theta ) = \operatornamewithlimits{arg\,max}_{p\in \mathcal {S}} \min _{j\ne I^*} \frac{ \left(\mu (\theta , I^*, w) - \mu (\theta , j, w)\right)^2 }{2 \Vert  X_{\rm pop}\Vert _{A}^2 \left(p_i^{-1} + p_j^{-1}\right)}.\) 
